4.12: Applications Using Direct Variation
Did you know that to find the flow rate of water in a pipeline, you can use the formula \begin{align*}Q=AV\end{align*}, where \begin{align*}Q\end{align*} is the flow rate, \begin{align*}A\end{align*} is the area of a cross-section of the pipe, and \begin{align*}V\end{align*} is the velocity of the water? This means that if you know the flow rate and the area, you can find the velocity, if you know the flow rate and the velocity, you can find the area, and if you know the area and the velocity, you can find the flow rate. In this Concept, you'll work with equations like this one to solve real-world problems involving direct variation.
Guidance
Direct variation has numerous real-world examples. You have already seen three examples: the area of a square is directly proportional to its side length; the distance you travel varies directly as the time you have been driving; and the total cost is directly proportional to the number of pounds of strawberries you purchase.
Newton's Second Law
In 1687, Sir Isaac Newton published the famous Principea Mathematica. It contained his Second Law of Motion. This law is often written as: \begin{align*}F=m \cdot a\end{align*}, where \begin{align*}F=\end{align*} the amount of force applied to an object with mass \begin{align*}(m)\end{align*} and \begin{align*}a=\end{align*} acceleration.
Acceleration is given in the units \begin{align*}\text{meters/second}^2\end{align*} and force is given in units of Newtons.
Example A
If a 175 Newton force causes a heavily loaded shopping cart to accelerate down the aisle with an acceleration of \begin{align*}2.5\ m/s^2\end{align*}, calculate the mass of the shopping cart.
Solution: This question is basically asking us to solve for the constant of proportionality. Let us compare the two formulas.
\begin{align*}&y=k \cdot x && \text{The direct variation equation}\\ &F = m \cdot a && \text{Newton's Second law}\end{align*}
We see that the two equations have the same form. The variable \begin{align*}y\end{align*} is equal to force and \begin{align*}x\end{align*} is equal to acceleration.
\begin{align*}175=m \cdot 2.5\end{align*}
Now solve for \begin{align*}m\end{align*}, the constant of variation.
\begin{align*}\frac{175}{2.5}& =\frac{m \cdot 2.5}{2.5}\\ m& =70\end{align*}
Example B
If a 175 Newton force causes a heavily-loaded shopping cart to accelerate down the aisle with an acceleration of \begin{align*}2.5\ m/s^2\end{align*}, calculate the force needed to accelerate the same cart at \begin{align*}6 \ m/s^2\end{align*}.
Solution: Now you know the constant of variation is 70. In this formula, 70 represents the mass. To find the force needed to move the cart at an accelerated rate of 6 meters/second, substitute 6 for \begin{align*}a\end{align*} and evaluate the equation.
When \begin{align*}a=6\end{align*}, \begin{align*}F=70 \cdot 6=420\end{align*}.
The force needed to accelerate the cart is 420 Newtons.
Solving Direct Variation Using Proportions
You can use the Cross Products Theorem of proportions to solve direct variation situations. Because the fraction \begin{align*}\frac{rise}{run}\end{align*} is constant in a direct variation situation, you can create a proportion.
Example C
Ohm’s Law states that the voltage \begin{align*}(V)\end{align*} is equal to the electrical current \begin{align*}(I)\end{align*} in amps times the resistance \begin{align*}(R)\end{align*} in ohms. Translating this to an equation, \begin{align*}V=I \cdot R\end{align*}.
Suppose an electronics device passed a current of 1.3 amps at a voltage of 2.6 volts. What was the current when the voltage was increased to 12 volts?
Ohm’s Law matches with the definition of direct variation, so you can write a proportion. \begin{align*}\frac{2.6 \ volts}{1.3 \ amps}=\frac{12 \ volts}{I \ amps}\end{align*}. Using the Cross Products Theorem, solve for \begin{align*}I\end{align*}. \begin{align*}I=6\end{align*}. Therefore, when the voltage was increased to 12 volts, the electronics device passed a current of 6 amps.
Guided Practice
Suppose a car uses 5 liters of gasoline to drive 40 kilometers. How much gasoline will the car require in order to drive 128 kilometers?
Solution:
This is a problem of direct variation involving a constant that is kilometers per liter of gasoline. You can use a proportion to solve this problem. When setting up proportions, you need to make sure that the same unit is on the same side: top, bottom, left, or right. The same units cannot be placed so that they are on a diagonal from each other. It's easiest if you set up your proportion so that the unknown value is not in the denominator of the fraction. Since we are solving for liters, we will place liters in the numerators.
\begin{align*}\text{Set up the proportion so that liters are both on top.} && \frac{5\text{ liters}}{40\text{ km}}&=\frac{x\text{ liters}}{128\text{ km}}\\ \text{Isolate the variable.} && 128 \cdot \frac{5}{40}&=x \\ \text{Simplify.} && 16&=x\end{align*}
Practice
Sample explanations for some of the practice exercises below are available by viewing the following video. Note that there is not always a match between the number of the practice exercise in the video and the number of the practice exercise listed in the following exercise set. However, the practice exercise is the same in both. CK-12 Basic Algebra: Direct Variation Models (11:11)
- What two methods can be used to solve a direct variation problem?
- True or false? Every linear equation is a direct variation situation.
In 3 – 7, determine the constant of variation in each exercise.
- \begin{align*}y\end{align*} varies directly as \begin{align*}x\end{align*}; when \begin{align*}x=4,y=48\end{align*}
- \begin{align*}d\end{align*} varies directly as \begin{align*}t\end{align*}; when \begin{align*}t=7,d=329\end{align*}
- \begin{align*}l\end{align*} varies directly as \begin{align*}h\end{align*}; when \begin{align*}l=112\end{align*}, \begin{align*}h=-16\end{align*}
- \begin{align*}m\end{align*} is directly proportional to \begin{align*}h\end{align*}; when \begin{align*}m=461.50\end{align*}, \begin{align*}h=89.6\end{align*}
- \begin{align*}z\end{align*} is directly proportional to \begin{align*}r;\end{align*} when \begin{align*}r=412,z=51.5\end{align*}
- Determine the equation of the strawberry purchase in the opener of this Concept.
- Dasan’s mom takes him to the video arcade for his birthday. In the first 10 minutes, he spends $3.50 playing games. If his allowance for the day is $20.00, how long can he keep playing games before his money is gone?
- The current standard for low-flow showerheads is 2.5 gallons per minute. Calculate how long it would take to fill a 30-gallon bathtub using such a showerhead to supply the water.
- An electrical device passes a force of 288 volts at 32 amps. Using Ohm’s Law, determine:
- The constant of proportionality
- The force needed to pass 65 amps
- The diameter of a circle is directly proportional to its radius. If a circle with a 2-inch diameter has a circumference of approximately 6.28 inches, what is the circumference of a 15-inch circle?
- Amin is using a hose to fill his new swimming pool for the first time. He starts the hose at 10:00 P.M. and leaves it running all night. At 6:00 A.M. he measures the depth and calculates that the pool is four-sevenths full. At what time will his new pool be full?
- Land in Wisconsin is for sale to property investors. A 232-acre lot is listed for sale for $200,500. Assuming the same price per acre, how much would a 60-acre lot sell for?
- The force \begin{align*}(F)\end{align*} needed to stretch a spring by a distance \begin{align*}x\end{align*} is given by the equation \begin{align*}F=k \cdot x\end{align*}, where \begin{align*}k\end{align*} is the spring constant (measured in Newtons per centimeter, N/cm). If a 12-Newton force stretches a certain spring by 10 cm, calculate:
- The spring constant, \begin{align*}k\end{align*}
- The force needed to stretch the spring by 7 cm
- The distance the spring would stretch with a 23-Newton force
- Determine the equations of graphs \begin{align*}a - d\end{align*} below.
Image Attributions
Here you'll use various formulas to solve real-world direct variation problems.